AME 436 Energy and Propulsion

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AME 436Energy and Propulsion

• Lecture 3
• Chemical thermodynamics concluded
• Equilibrium thermochemistry

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Outline

• Why do we need to invoke chemical equilibrium?
• Degrees Of Reaction Freedom (DORFs)
• Conservation of atoms
• Second Law of Thermodynamics for reactive systems
• Equilbrium constants
• Application of chemical equilibrium to
  hydrocarbon-air combustion
• Application of chemical equilibrium to
  compression/expansion

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Why do we need chemical equilibrium?

• (From lecture 2) What if we assume more products,
  e.g.
• 1CH4 2(O2 3.77N2) ? ? CO2 ? H2O ? N2 ?
  CO
• In this case how do we know the amount of CO vs.
  CO2?
• And if if we assume only 3 products, how do we
  know that the preferred products are CO2, H2O
  and N2 rather than (e.g.) CO, H2O2 and N2O?
• Need chemical equilibrium to decide - use 2nd Law
  to determine the worst possible end state of the
  mixture

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Degrees of reaction freedom (DoRFs)

• If we have a reacting soup of CO, O2, CO2, H2O,
  H2 and OH, can we specify changes in the amount
  of each molecule independently? No, we must
  conserve each type of atom (3 in this case)
• Conservation of
• C atoms nCO nCO2 constant
• O atoms nCO 2nCO2 2nO2 nH2O nOH
  constant
• H atoms 2nH2O 2nH2 nOH constant
• 3 equations, 6 unknown nis ? 3 degrees of
  reaction freedom (DoRFs)
• of DoRFs of different molecules (n) - of
  different elements
• Each DoRF will result in the requirement for one
  equilibrium constraint

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Conservation of atoms

• Typically we apply conservation of atoms by
  requiring that the ratios of atoms are constant
• C atoms nCO nCO2 constant
• O atoms nCO 2nCO2 2nO2 nH2O nOH
  constant
• H atoms 2nH2O 2nH2 nOH constant
• Specifying nO/nH also would be redundant, so the
  number of atom ratio constraints of atoms - 1
• What are these constants? Depends on initial
  mixture, e.g. for stoichiometric CH4 in O2, nC/nO
  1/4, nC/nH 1/4

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2nd law of thermo for reacting systems

• Constraints for reacting system
• First law dE ?Q - ?W ?Q - PdV
• Second law dS ?Q/T
• Combine TdS - dU - PdV 0 for any allowable
  change in the state of the system
• For a system at fixed T and P (e.g. material in a
  piston-cylinder with fixed weight on top of
  piston, all placed in isothermal bath
  d(TS-U-PV) 0, or per unit mass d(Ts-u-Pv) 0
• Define Gibbs function g ? h - Ts u Pv - Ts
• Thus for system at fixed T and P d(-g) 0 or
  dg 0
• Thus at equilibrium, dg 0 (g is minimum)
• Similarly, for system at fixed U and V (e.g.
  insulated chamber of fixed volume) ds 0, at
  equilibrium ds 0 (s is maximum)

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2nd law of thermo for reacting systems

• For a mixture of ideal gases with 1 DoRF that
  reacts according to ?AA ?BB ? ?CC ?DD , e.g.
• 1 CO2 ? 1 CO .5 O2
• A CO2, ?A 1, B nothing, ?B 0, C CO, ?C
  1, D O2, ?D 0.5
• with dg 0, it can be shown

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Equilibrium constants

• Examples of tabulated data on K - (double-click
  table to open Excel spreadsheet with all data for
  CO, O, CO2, C, O2, H, OH, H2O, H2, N2, NO at 200K
  - 6000K)

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Chemical equlibrium - example

• For a mixture of CO, O2 and CO2 (and nothing
  else, note 1 DoRF for this case) at 10 atm and
  2500K with CO 12, what are the mole fractions
  of CO, O2 and CO2?
• 1 CO2 ? 1 CO .5 O2
• 3 equations, 3 unknowns XCO2 0.9097, XO2
  0.0301, XCO 0.0602
• At 10 atm, 1000K XCO2 1.0000, XO2 4.48 x
  10-8, XCO 8.96 x 10-8
• At 1 atm, 2500K XCO2 0.8186, XO2 0.0605, XCO
  0.1209
• With N2 addition, CON 127, 10 atm, 2500K
  XCO2 0.1980, XO2 0.0109, XCO 0.0218, XN2
  0.7693 (XCO2/XCO 9.1 vs. 15.1 without N2
  dilution) (note still 1 DoRF in this case)
• Note high T, low P and dilution favor dissociation

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Chemical equlibrium - hydrocarbons

• Reactants CxHy rO2 sN2 (not necessarily
  stoichiometric)
• Assumed products CO2, CO, O2, O, H2O, OH, H,
  H2, N2, NO
• How many equations?
• 10 species, 4 elements ? 6 DoRFs ? 6 equil.
  constraints
• 4 types of atoms ? 3 atom ratio constraints
• Conservation of energy (hreactants hproducts)
  (constant pressure reaction) or ureactants
  uproducts (constant volume reaction)
• Pressure constant or (for const. vol.)
• 6 3 1 1 1 12 equations
• How many unknowns?
• 10 species ? 10 mole fractions (Xi)
• Temperature
• Pressure
• 10 1 1 12 equations

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Chemical equlibrium - hydrocarbons

• Equilibrium constraints - not a unique set, but
  for any set
• Each species appear in at least one constraint
• Each constraint must have exactly 1 DoRF
• Example set

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Chemical equlibrium - hydrocarbons

• Atom ratios
• Sum of all mole fractions 1
• Conservation of energy (constant P shown)

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Chemical equlibrium - hydrocarbons

• This set of 12 simultaneous nonlinear algebraic
  equations looks hopeless, but computer programs
  (using slightly different methods more amenable
  to automation) (e.g. GASEQ) exist
• Typical result, for stoichiometric CH4-air, 1
  atm, constant P

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Chemical equlibrium - hydrocarbons

• Most of products are CO2, H2O and N2 but some
  dissociation (into the other 7 species) occurs
• Product ? is much lower than reactants - affects
  estimation of compression / expansion processes
  using Pv? relations
• Bad things like NO and CO appear in relatively
  high concentrations in products, but will
  recombine to some extent during expansion
• By the time the expansion is completed, according
  to equilibrium calculations, practically all of
  the NO and CO should disappear, but in reality
  they dont - as T and P decrease during
  expansion, reaction rates decrease, thus at some
  point the reaction becomes frozen, leaving NO
  and CO stuck at concentrations MUCH higher than
  equilibrium

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Adiabatic flame temp. - hydrocarbons
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Adiabatic flame temp - hydrocarbons

• Adiabatic flame temperature (Tad) peaks slightly
  rich of stoichiometric - since O2 is highly
  diluted with N2, burning slightly rich ensures
  all of O2 is consumed without adding a lot of
  extra unburnable molecules
• Tad peaks at 2200K for CH4, slightly higher for
  C3H8, iso-octane (C8H18) practically
  indistinguishable from C3H8
• H2 has far heating value per unit fuel mass, but
  only slightly higher per unit total mass (due to
  heavy air), so Tad not that much higher
• Also - massive dissociation as T increases above
  2400K, keeps peak temperature down near
  stoichiometric
• Also - since stoichiometric is already 29.6 H2
  in air (vs. 9.52 for CH4, 4.03 for C3H8), so
  going rich does not add as many excess fuel
  molecules
• CH4 - O2 MUCH higher - no N2 to soak up thermal
  energy without contributing heat release
• Constant volume - same trends but higher Tad

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Compression / expansion

• As previously advertised, compression / expansion
  are assumed to occur at constant entropy (not
  constant h or u)
• Three levels of approximation
• Frozen composition (no change in Xis),
  corresponds to infinitely slow reaction
• Equilibrium composition (Xis change to new
  equilibrium), corresponds to infinitely fast
  reaction (since once we get to equilibrium, no
  further change in composition can occur)
• Reacting composition (finite reaction rate, not
  infinitely fast or slow) - more like reality but
  MUCH more difficult to analyze since rate
  equations for 100s or 1000s of reactions are
  involved
• Which gives best performance?
• Equilibrium - youre getting everything the gas
  has to offer recombination (e.g. H OH ? H2O)
  gives extra heat release, thus more push on
  piston or more kinetic energy of exhaust
• Frozen - no recombination, no extra heat release
• Reacting - somewhere between, most realistic

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Compression / expansion

• Entropy of an ideal gas - depends on P as well as
  T (unlike h and u, which depend ONLY on T)
• entropy of species i at reference
  pressure (1 atm) and temperature T (not
  necessarily 298K) (see my tables)
• ?ln(P/Pref) - entropy associated with pressure
  different from 1 atm -
• P gt Pref leads to decrease in s
• ?Xiln(Xi) - entropy associated with mixing (Xi lt
  1 means more than 1 specie is present - always
  leads to increase in entropy
• since -Xiln(Xi) gt 0)
• Denominator is just the average molecular weight
• Units of s are J/kgK
• Use sreactants sproducts constant for
  compression / expansion instead of hreactants
  hproducts or ureactants uproducts
• All other relations (atom ratios, ?Xi 1,
  equilibrium constraints) still apply

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Compression / expansion

• Example - expansion of CO2-O2-CO mixture from 10
  atm, 2500K to 1 atm in steady-flow control volume
  (e.g. nozzle) or control mass (e.g
  piston/cylinder)
• Initial state
• XCO2 0.9097, XO2 0.0301, XCO 0.0602,
• h -5782.92 kJ/kg, u -6269.88 kJ/kg, s
  7073.83 J/kgK
• Example calculation of s At 2500K

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Compression / expansion

• Expand at constant entropy to 1 atm, frozen
  composition
• T 1803K, XCO2 0.9097, XO2 0.0301, XCO
  0.0602,
• h -6740.38 kJ/kg, u -7071.57 kJ/kg, s
  7073.73 J/kgK
• Work done (control volume, steady flow) hbefore
  - hafter
• 957.51 kJ/kg
• Work done (control mass) ubefore - uafter
  821.74 kJ/kg
• Expand at constant entropy to 1 atm, equilibrium
  composition
• T 1996K, XCO2 0.9783, XO2 0.0073, XCO
  0.0145 (significant recombination), h -6778.02
  kJ/kg, u -7157.88 kJ/kg,
• s 7073.83 J/kgK
• Work done (control volume, steady flow) 995.12
  kJ/kg
• (3.9 higher)
• Work done (control mass) 888.05 kJ/kg (8.1
  higher)
• Moral let your molecules recombine!

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Summary - Lecture 3

• In order to understand what happens to a
  chemically reacting mixture if we wait a very
  long time, we need to apply
• 1st Law of Thermodynamics (conservation of
  energy) - but this doesnt tell us what the
  allowable direction of the reaction is A ? B or
  B ? A are equally valid according to the 1st Law
• 2nd Law of Thermodynamics (increasing entropy) -
  invokes restrictions on the direction of
  allowable processes (if A ? B is allowed then B ?
  A isnt)
• Equilibrium occurs when the worst possible end
  state is reached once this point is reached, no
  further chemical reaction is possible unless
  something is changed, e.g. T, P, V, etc.
• The statements of the 2nd law are
• ds 0 (constant u v, e.g. a rigid, insulated
  box)
• dg d(h - Ts) 0 (constant T and P, e.g.
  isothermal piston/cylinder)
• The application of the 2nd law leads to
  complicated looking expressions called
  equilibrium constraints that involve the
  concentrations of each type of molecule present
  in the mixture
• These equilibrium constraints are coupled with
  conservation of energy, conservation of each type
  of atom, and the ideal gas law (or other
  equations of state, not discussed in this class)
  to obtain a complete set of equations that
  determine the end state of the system, e.g. the
  combustion products